Question

What is the percentage of oxygen in C6H1206?​

Answer 1

number of oxygen : 6

number of whole atoms in molchol :

6 + 12 + 6 = 24

Percentage of oxygen = n o o / n o a i m × 100

P o o = 6 / 24 × 100

P o o = 1/4 × 100

P o o = % 25

Answer 2

`\huge{ \mathrm{  \underline{ Answer} \:  \:  ✓ }}`

Mass of elements in
`\mathrm{C_6H_(12)O_6}` are :

• O =
  `16 u`

• C =
  `12 u`

• H =
  `1 u`

Total mass of
`\mathrm{C_6H_(12)O_6}` is :

• 
  `(12u * 6) + (1u * 12) +( 16u * 6)`

• 
  `72u + 12u + 96u`

• 
  `180u`

Total Mass of Oxygen in
`\mathrm{C_6H_(12)O_6}` is :

• 
  `16u * 6`

• 
  `96u`

Percentage of Oxygen by mass in
`\mathrm{C_6H_(12)O_6}` is :

`\boxed{ (total \: \: mass \: \: of \: \: oxygen)/(total \: \: mass \: of \: glucose ) * 100 }`

• 
  `(96u)/(180u) * 100`

• 
  `(32u)/(3u) * 5`

• 
  `(160u)/(3u)`

• 
  `53.33 \: \%`

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